Is $\phi(U)$ a Subspace of $\mathbb{R}^m$?

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In summary, the author is discussing how to find a subspace of a larger space. They state that the linear map $\phi$ is subspace and that there is no inverse. They also state that $\phi^{-1}$ is not a linear map. They ask for help with their problem and the user provides a solution.
  • #1
mathmari
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Hey! :eek:

Let $1\leq m, n\in \mathbb{N}$, let $\phi :\mathbb{R}^n\rightarrow \mathbb{R}^m$ a linear map and let $U\leq_{\mathbb{R}}\mathbb{R}^n$, $W\leq_{\mathbb{R}}\mathbb{R}^m$ be subspaces.
I want to show that:
  1. $\phi (U)$ is subspace of $\mathbb{R}^m$.
  2. $\phi^{-1} (W)$ is subspace of $\mathbb{R}^n$.
I have done the following:

  1. We have that $\phi (U)=\{\phi (u) \mid u\in U\}$.

    Since $U$ is a subspace we have that $0\in U$. Therefore $\phi (0)\in \phi (U)$. Since $\phi$ is a linear map we have that $\phi (0)=0$ and so we get that $0\in \phi (U)$.

    Let $\phi (u_1), \phi (u_2)\in \phi (U)$. Then we have that $\phi (u_1)+\phi (u_2)=\phi (u_1+u_2)$, since $\phi$ is linear.
    Since $U$ is a subspace we have that since $u_1, u_2\in U$ then $u_1+u_2\in U$. Therefore we get that $\phi (u_1+u_2)\in \phi (U)$ and so we have that $\phi (u_1)+\phi (u_2)\in \phi (U)$.

    Let $\lambda\in \mathbb{R}$ and $\phi (u_1)\in \phi (U)$. Then we have that $\lambda \phi (u_1)=\phi (\lambda u_1)$, since $\phi$ is linear.
    Since $U$ is a subspace we have that since $\lambda\in \mathbb{R}$ and $u_1\in U$ then $\lambda u_1\in U$. Therefore we get that $\phi (\lambda u_1)\in \phi (U)$ and so we have that $\lambda \phi (u_1)\in \phi (U)$.



    That means that $\phi (U)$ is subspace of $\mathbb{R}^m$.
    Is everything correct? (Wondering)
  2. We have that $\phi$ is linear. Does it follow then that $\phi^{-1}$ is also linear? (Wondering)
 
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  • #2
mathmari said:
Is everything correct?

Hey mathmari!

Yep. (Nod)

mathmari said:
2. We have that $\phi$ is linear. Does it follow then that $\phi^{-1}$ is also linear?

Let's see... suppose we pick $\phi: u \mapsto 0$.
That is a linear map isn't it?
What is $\phi^{-1}$? Is it a linear map? Is it a function for that matter? (Wondering)
 
  • #3
Klaas van Aarsen said:
Let's see... suppose we pick $\phi: u \mapsto 0$.
That is a linear map isn't it?
What is $\phi^{-1}$? Is it a linear map? Is it a function for that matter? (Wondering)

There is no inverse, is there? (Wondering)
 
  • #4
mathmari said:
There is no inverse, is there?

Indeed. So we'll have to solve the problem differently. (Thinking)
 
  • #5
Klaas van Aarsen said:
Indeed. So we'll have to solve the problem differently. (Thinking)

But how? (Wondering)
 
  • #6
mathmari said:
But how? (Wondering)

As a linear map, $\phi(0)=0$, so $0\in\phi^{-1}(\{0\})$.
Since W is a subspace, it must contain 0.
So we must have that $0\in \phi^{-1}(W)$, don't we? (Wondering)

Suppose $u,v\in \phi^{-1}(W)$, then what can we find out about $u+v$? (Wondering)
 
  • #7
Klaas van Aarsen said:
Suppose $u,v\in \phi^{-1}(W)$, then what can we find out about $u+v$? (Wondering)

We have that $\phi(u), \phi(v) \in W$. Since $W$ is a subspace we have that $\phi(u) +\phi (v) \in W$. Since $\phi$ is linear we get that $\phi (u+v) \in W$. We apply the inverse and we get $u+v\in \phi^{-1}(W)$.

Is everything correct? (Wondering)
 
  • #8
mathmari said:
We have that $\phi(u), \phi(v) \in W$. Since $W$ is a subspace we have that $\phi(u) +\phi (v) \in W$. Since $\phi$ is linear we get that $\phi (u+v) \in W$. We apply the inverse and we get $u+v\in \phi^{-1}(W)$.

Is everything correct? (Wondering)

Yep. (Nod)
 
  • #9
Klaas van Aarsen said:
Yep. (Nod)

I thought about that again and now I got stuck. I said that $\phi (u), \phi (v)\in W$. Is this correct? Does this hold because $W$ is a subspace of $\mathbb{R}^m$ and the image of $\phi$ is $\mathbb{R}^m$ ? (Wondering)
 
  • #10
mathmari said:
I thought about that again and now I got stuck. I said that $\phi (u), \phi (v)\in W$. Is this correct? Does this hold because $W$ is a subspace of $\mathbb{R}^m$ and the image of $\phi$ is $\mathbb{R}^m$ ?

We assumed that $u\in\phi^{-1}(W)$. Doesn't this mean by definition that $\phi(u)\in W$?
That is, isn't $\phi^{-1}(W)\overset{\text{def}}{=}\{x : \phi(x) \in W\}$? (Wondering)
The same applies to $v$.
 
  • #11
Klaas van Aarsen said:
We assumed that $u\in\phi^{-1}(W)$. Doesn't this mean by definition that $\phi(u)\in W$?
That is, isn't $\phi^{-1}(W)\overset{\text{def}}{=}\{x : \phi(x) \in W\}$? (Wondering)
The same applies to $v$.

Ahh yes! (Blush)

Thank you very much! (Yes)
 

FAQ: Is $\phi(U)$ a Subspace of $\mathbb{R}^m$?

What is a subspace?

A subspace is a subset of a vector space that satisfies the three properties of closure under vector addition, closure under scalar multiplication, and contains the zero vector.

How do you show that a set is a subspace?

To show that a set is a subspace, you must prove that it satisfies the three properties of closure under vector addition, closure under scalar multiplication, and contains the zero vector. This can be done by using mathematical proofs or by showing that the set is a span of a set of vectors.

Why is it important to show that a set is a subspace?

It is important to show that a set is a subspace because it allows us to determine if the set is a vector space. This is useful in many areas of mathematics, such as linear algebra and functional analysis, where vector spaces are commonly used.

Can a subspace be empty?

Yes, a subspace can be empty. However, an empty set is only considered a subspace if it contains the zero vector. If the empty set does not contain the zero vector, it is not considered a subspace.

What are some common examples of subspaces?

Some common examples of subspaces include the x-y plane in three-dimensional space, the set of all polynomials of degree n or less, and the set of all real-valued continuous functions on a closed interval. However, any subset of a vector space that satisfies the three properties of a subspace can be considered a subspace.

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